A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th

Response:

00100111

Explanation:

We have been given;

10010110

10010000

Add them following standard binary addition rules

10010110

10010000

-------------

(1)00100110

-------------

ignore the leading (1) because it is a carry.

Increase the result by 1 to achieve a 1's complement sum

00100110 + 1 = 00100111

Final Result: 00100111

Response:

1. To find the volume of the glass shell (Vg), simply subtract the volume of the empty part of the jar (Ve) from the total volume of the jar (Vj):

Vg = Vj - Ve

Volume can be calculated by multiplying the base (B) with the height (h). The base of the jar is a circle, thus its area is πr^2 (where r indicates the radius).

The radius differs based on the jar's section: the inner radius for the empty part is d = 3 in, while for the total jar it includes the glass thickness a = 3 + 3/16 = 3.1875 in.

The height of the entire jar is given as h = 6 in, whereas for the empty portion, it's the total height minus the thickness of the glass h' = 6 - 0.1875 = 5.8125 in.

Now we can perform the calculations:

Vj = πa^2 • h = 191.42 in^3

Ve = πd^2 • h' = 164.26 in^3

Thus, the volume of the glass shell equals Vj - Ve, resulting in 27.16 in^3.

2. The mass of the glass jar can be determined by multiplying the density of the glass with the volume:

m = ρ • Vg

The glass density is provided in cubic feet, so we first convert it to cubic inches by dividing by 1728:

ρ = 165 lb/ft^3 / 1728 = 0.095 lb/in^3

m = 0.095 lb/in^3 • 27.16 in^3 = 2.59 lb

5. To calculate the weight and volume of the displaced water, we first need to ascertain how deep the jar sinks (H), as the volume of displaced water equals the submerged volume of the jar. The jar will descend until the gravitational force downwards equals the buoyancy force upwards. The displaced water volume is πa^2 • H, and the buoyancy is calculated as ρw • g • Vd (where ρw is the density of water, defined as 62.5 lb/ft^3 / 1728 = 0.036 lb/in^3, and Vd is the displaced water volume).

Thus, the buoyancy can be represented as:

B = ρw • g • πa^2 • H

Setting buoyancy equal to gravity:

B = m • g (where m is the mass of the jar). Therefore, we have:

ρw • g • πa^2 • H = m • g

From this, simplifying gives:

ρw • πa^2 • H = m

We can derive H:

H = m / (ρw • πa^2)

H = 2.25 inches

This indicates the jar will sink 2.25 inches into the water.

3. Calculating the volume of displaced water is straightforward. It matches the volume of the submerged jar:

Vd = πa^2 • H

Vd = 71.94 in^3

4. Lastly, to determine the weight of the displaced water:

m = ρw • Vd

m = 0.036 lb/in^3 • 71.94 in^3

m = 2.59 lb

As evident, the mass of the jar aligns with the mass of the displaced water. Following this logic could have simplified our calculations, but I chose to elaborate for clarity.

Explanation: For each kilogram of Pu-239 decaying, there is a decay heat release of 1.9 Watts. To find the heat released by 1.25 moles of Pu-239 over a duration of 3 hours (in calories): First, we find the mass of Pu-239 in 1.25 moles. The molar mass of Pu-239 is 244 g/mol or 0.244 kg/mol. Thus, 1.25 moles = 1.25 × 0.244 = 0.305 kg. Each kilogram yields 1.9 Watts, hence: 0.305 kg of Pu-239 results in (0.305 × 1.9) Watts; approximately 0.5795 Watts. Note that Watts equate to J/s. Using the relation Power = Energy/time, we have Energy = power × time. Knowing: Power = 0.5795 Watts, Time = 3 hours = 10800 seconds, we calculate: Energy = 0.5795 × 10800 = 6258.6 J. Finally, converting to calories: 6258.6 J equals 1495.81 calories when multiplied by 0.239, thus reaching the total of approximately 1495.81 calories.